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Nana76 [90]
3 years ago
7

Please help i have to pass

Physics
2 answers:
lana66690 [7]3 years ago
3 0
Hey kid!

So let’s not make this complicated. Obviously this word problem is wanting to confuse you with all these fancy words but, we’re not letting that get in the way!

Our two numbers are 6 and 12. If you think about it... 12 divided by 6 equals 2!!
That would make your answer B) 2m.

Happy to help!
~Brooke❤️
sertanlavr [38]3 years ago
3 0
Pe = 12 J

pe = mgh
solve for h

h = pe/mg where g = 9.81 m/s^2

h = 12 J / (6 kg × 9.81 m/s^2) = 0.20387 m

hence answer is A
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A parallel plate capacitor is constructed with a dielectric slab with κ = 1.5 inserted between the plates. The area of each plat
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Answer:

The value is V =  4.533  \  V

Explanation:

From the question we are told that

The dielectric constant is k = 1.5

The area of each plate is A =  5 \ cm^2 = 0.0005 m^2

The distance between the plates is d= 1 \  mm  =  0.001 \  m

The charge on the capacitor is Q =  6*10^{-11} \  C

Generally the electric field in a vacuum is mathematically represented as

E_o  =  \frac{V_o}{d}

Generally V_o is the voltage of the capacitor which is mathematically represented as

V_o =  \frac{Q}{C_o}

Here C_o is the capacitance of the capacitor in a vacuum which is mathematically represented as

C_o =  \frac{\epsilon_o * A}{d}

Here epsilon_o is a constant with value epsilon_o= 8.85*10^{-12} C^2 \cdot N^{-1} \cdot m^{-2}

=> C_o =  \frac{8.85*10^{-12} * 0.0005 }{0.001}

=> C_o =  4.425 *10^{-12}  \  F

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V_o =  \frac{6*10^{-11} }{  4.425 *10^{-12}}

V_o =  13.6 \  V

So

E_o  =  \frac{ 13.6}{0.001}

=> E_o  = 13600 \ V/m

The electric field when the dielectric slab is inserted is mathematically represented as

E =  \frac{E_o}{k}

=> E =  \frac{13600}{1.5}

=> E =9067  \ V/m

Generally the electric field between the plates is mathematically evaluated as

E_{N} =  E_o - E

=> E_{N} =  13600- 9067

=> E_{N} = 4533 \  V/m

Generally the potential difference between the plates is

V =  4533   *  0.001

=> V =  4.533  \  V

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